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A085697
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a(n) = T(n)^2, where T(n) = A000073(n) is the n-th tribonacci number.
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17
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0, 0, 1, 1, 4, 16, 49, 169, 576, 1936, 6561, 22201, 75076, 254016, 859329, 2907025, 9834496, 33269824, 112550881, 380757169, 1288092100, 4357584144, 14741602225, 49870482489, 168710633536, 570743986576, 1930813074369, 6531893843049
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OFFSET
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0,5
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COMMENTS
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In general, squaring the terms of a third-order linear recurrence with signature (x,y,z) will result in a sixth-order recurrence with signature (x^2 + y, x^2*y + z*x + y^2, x^3*z + 4*x*y*z - y^3 + 2*z^2, x^2*z^2 - x*y^2*z - z^2*y, z^2*y^2 - z^3*x, -z^4). - Gary Detlefs, Jan 10 2023
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REFERENCES
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R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.
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LINKS
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FORMULA
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G.f.: x^2*( 1-x-x^2-x^3 )/( (1-3*x-x^2-x^3)*(1+x+x^2-x^3) ).
a(n+6) = 2*a(n+5) + 3*a(n+4) + 6*a(n+3) - a(n+2) - a(n).
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MATHEMATICA
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LinearRecurrence[{2, 3, 6, -1, 0, -1}, {0, 0, 1, 1, 4, 16}, 30] (* Harvey P. Dale, Oct 26 2020 *)
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PROG
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(Maxima)
t[0]:0$ t[1]:0$ t[2]:1$
t[n]:=t[n-1]+t[n-2]+t[n-3]$
(Magma)
R<x>:=PowerSeriesRing(Integers(), 40); [0, 0] cat Coefficients(R!( x^2*(1-x-x^2-x^3)/((1-3*x-x^2-x^3)*(1+x+x^2-x^3)) )); // G. C. Greubel, Nov 20 2021
(Sage)
@CachedFunction
if (n<2): return 0
elif (n==2): return 1
else: return T(n-1) +T(n-2) +T(n-3)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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